
/************************************************************
 *
 * Cube of any integer can be written as the sum of squares
 * of two integers, such as 1985^3=19711052-19691202. Input
 * a integer N, then output it in form of N^3=X^2-Y^2.
 *
 * n = [(x-y)/n] . [(x+y)/n]
 * (x-y)/n = k1
 * (x+y)/n =k2
 * resolve it and we get
 * x = (k2+k1)n / 2
 * y = (k2-k1)n / 2
 * if we take proper k1 and k2, s.t. k1.k2=n
 * we can get the values of x and y.
 * In particular, we can make k1=1, k2=n. Thus, we get
 * x = n(n+1) / 2
 * y = n(n-1) / 2
 *
 * Date: 5/28/2008
 ***********************************************************/

int main()
{
	return 0;
}
